# If one number is thrice the other and their sum is $16$, find the numbers

If one number is thrice the other and their sum is $16$, find the numbers.

I tried, Let the first number be $x$ and the second number be $y$ Acc. to question

\begin{align} x&=3y &\iff x-3y=0 &&(1)\\ x&=16-3y&&&(2) \end{align}

• One number is $x$, the other $3x$. They add to 16, so ... . – Chris Leary Jul 27 '14 at 20:01
• the second part of the questions states "their sum is 16". this can be expressed as x+y=16 – Mufasa Jul 27 '14 at 20:01

The problem statement says $$x+3x=16,$$ hence $$x=4,\\3x=12.$$

I assume you have $$x=\color{red}{3y}, ~~x+y=16$$ Then $3(x+\color{red}{y})=3\times 16=48$ and so $3x+\color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...

$$x=3y$$ $$y+x=16 \Rightarrow y+3y=16 \Rightarrow 4y=16 \Rightarrow y=4$$

So, $x=12$.

Let the 1st number be $$x$$ and the 2nd number be $$3x$$. Since $$x + 3x = 16$$, $$4x = 16$$, so $$x=4$$. Therefore, 1st number is $$4$$ and the other is $$12$$.

Let the first number be $$x$$.

Let the second number be $$y$$.

According to question

$$\tag{1} x+y=16$$ $$\tag{2} x=3y$$ So, $$x-3y=0 \tag{2}$$ Multiply equation $$(1)$$ by $$3$$.

Solve both equations:

$$\tag{1} 3x+3y=48$$ $$\tag{2} x-3y=0$$ $$\tag{1) + (2}4x=48$$ $$\tag{3}x=12$$ Putting in equation $$(1)$$: $$\tag{1} x+y=16$$ $$\tag{1),(3} 12+y=16$$ $$\tag{4}y=16-12$$ $$\tag{5}y=4$$

• Welcome to MSE! Please use MathJax so the others can better benefit from your questions and answers. – Bill O'Haran May 3 '18 at 15:57

Let the first number be $$x$$ And second which is thrice be $$3y$$

Acc to ques..

$$x=3y\ldots \textrm{equation 1}$$

$$x+y=16\ldots\textrm{equation 2}$$

By Elimination method

       x-3y  =    0
-(x+ y) =  -16
______________
0- 4y = - 16


Dividing the equation by $$(-4)$$. The result is $$y=4$$

Put $$y=4$$ into the first equation :$$x=3\cdot 4\Rightarrow x=12$$